You can also use df.icol(n)
to access a column by integer.
Update: icol
is deprecated and the same functionality can be achieved by:
df.iloc[:, n] # to access the column at the nth position
The method .transpose() converts columns to rows and rows to column, hence you could even write
df.transpose().ix[3]
You could use label based using .loc or index based using .iloc method to do column-slicing including column ranges:
In [50]: import pandas as pd
In [51]: import numpy as np
In [52]: df = pd.DataFrame(np.random.rand(4,4), columns = list('abcd'))
In [53]: df
Out[53]:
a b c d
0 0.806811 0.187630 0.978159 0.317261
1 0.738792 0.862661 0.580592 0.010177
2 0.224633 0.342579 0.214512 0.375147
3 0.875262 0.151867 0.071244 0.893735
In [54]: df.loc[:, ["a", "b", "d"]] ### Selective columns based slicing
Out[54]:
a b d
0 0.806811 0.187630 0.317261
1 0.738792 0.862661 0.010177
2 0.224633 0.342579 0.375147
3 0.875262 0.151867 0.893735
In [55]: df.loc[:, "a":"c"] ### Selective label based column ranges slicing
Out[55]:
a b c
0 0.806811 0.187630 0.978159
1 0.738792 0.862661 0.580592
2 0.224633 0.342579 0.214512
3 0.875262 0.151867 0.071244
In [56]: df.iloc[:, 0:3] ### Selective index based column ranges slicing
Out[56]:
a b c
0 0.806811 0.187630 0.978159
1 0.738792 0.862661 0.580592
2 0.224633 0.342579 0.214512
3 0.875262 0.151867 0.071244
You can access multiple columns by passing a list of column indices to dataFrame.ix.
For example:
>>> df = pandas.DataFrame({
'a': np.random.rand(5),
'b': np.random.rand(5),
'c': np.random.rand(5),
'd': np.random.rand(5)
})
>>> df
a b c d
0 0.705718 0.414073 0.007040 0.889579
1 0.198005 0.520747 0.827818 0.366271
2 0.974552 0.667484 0.056246 0.524306
3 0.512126 0.775926 0.837896 0.955200
4 0.793203 0.686405 0.401596 0.544421
>>> df.ix[:,[1,3]]
b d
0 0.414073 0.889579
1 0.520747 0.366271
2 0.667484 0.524306
3 0.775926 0.955200
4 0.686405 0.544421
Source: Stackoverflow.com